Calculate the second ionization energy of the metal M (AH¡on2 in kJ/mol) using the following data: Lattice enthalpy of MO(s), AH = -2278 kJ/mol Bond dissociation enthalpy of O2(g) = +498 kJ/mol First electron affinity of O = -141 kJ/mol Second electron affinity of O = +744 kJ/mol Enthalpy of sublimation of M = + 125 kJ/mol First ionization energy of M = + 309 kJ/mol Standard enthalpy of formation of MO(s), AH = -341 kJ/mol

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Calculate the second ionization energy of the metal M (AH¡on2 in kJ/mol) using the
following data:
Lattice enthalpy of MO(s), AH = -2278 kJ/mol
Bond dissociation enthalpy of O2(g) = +498 kJ/mol
%3D
First electron affinity of O = -141 kJ/mol
Second electron affinity of O = +744 kJ/mol
Enthalpy of sublimation of M = + 125 kJ/mol
First ionization energy of M = + 309 kJ/mol
Standard enthalpy of formation of MO(s), AHf
= -341 kJ/mol
Transcribed Image Text:Calculate the second ionization energy of the metal M (AH¡on2 in kJ/mol) using the following data: Lattice enthalpy of MO(s), AH = -2278 kJ/mol Bond dissociation enthalpy of O2(g) = +498 kJ/mol %3D First electron affinity of O = -141 kJ/mol Second electron affinity of O = +744 kJ/mol Enthalpy of sublimation of M = + 125 kJ/mol First ionization energy of M = + 309 kJ/mol Standard enthalpy of formation of MO(s), AHf = -341 kJ/mol
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